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BARSIC [14]
3 years ago
11

Interpret using 12 as a reference point how can you tell that the hands of the clock on the right have moved from the previous p

osition

Physics
1 answer:
zubka84 [21]3 years ago
6 0

Answer:

The answer is "6:45 and 7:15".

Explanation:

In the given attache file two clocks are added, that's explanation defined as follows:

In the left image, it may assume, that the clock only at the right-hand side of a scene is move-in its previous position, which is shown to the left, as it was at 6:45 on the left but in the right image it will display the time that is 7:15.

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suggest an experiment to prove that the rate of evaporation of a liquid depends on its surface area vapour already present in su
gulaghasi [49]
That's two different things it depends on:

-- surface area exposed to the air
AND
-- vapor already present in the surrounding air.

Here's what I have in mind for an experiment to show those two dependencies:

-- a closed box with a wall down the middle, separating it into two closed sections;

-- a little round hole in the east outer wall, another one in the west outer wall,
and another one in the wall between the sections;
So that if you wanted to, you could carefully stick a soda straw straight into one side,
through one section, through the wall, through the other section, and out the other wall.

-- a tiny fan that blows air through a tube into the hole in one outer wall.

<u>Experiment A:</u>

-- Pour 1 ounce of water into a narrow dish, with a small surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
-- Pour 1 ounce of water into a wide dish, with a large surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
<span><em>Show that the 1 ounce of water evaporated faster </em>
<em>when it had more surface area.</em></span>
============================================
============================================

<u>Experiment B:</u>

-- Again, pour 1 ounce of water into the wide dish with the large surface area.
-- Again, set the dish in the second half of the box ... the one the air passes
through just before it leaves the box.
-- This time, place another wide dish full of water in the <em>first section </em>of the box,
so that the air has to pass over it before it gets through the wall to the wide dish
in the second section.  Now, the air that's evaporating water from the dish in the
second section already has vapor in it before it does the job.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
==========================================
<em>Show that it took longer to evaporate when the air </em>
<em>blowing over it was already loaded with vapor.</em>
==========================================
6 0
3 years ago
15. A car is stationary at the top of a hill with the engine
White raven [17]

Answer:

kinetic energy

Explanation:

hop it is helpful

mark me brainlist

5 0
3 years ago
when an object is charged by contact, what kind of charge does the object have compared with the charge on the on the object giv
sveta [45]

Answer:

the charge that is given by the object is positive charge and the object which is taking the charge is negetively charged

Explanation:

5 0
3 years ago
I need help on putting this diagram in order.
morpeh [17]
In what type of order are you supposed to put it in?
3 0
3 years ago
A disk of mass M and radius R rotates at angular velocity ω0. Another disk of mass M and radius r is dropped on top of the rotat
AleksandrR [38]

Answer:

\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}

Explanation:

As we know that there is no external torque on the system of two disc

then the angular momentum of the system will remains conserved

So we will have

L_i = L_f

now we have

L_i = (\frac{1}{2}MR^2)\omega_o

also we have

L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega

now from above equation we have

(\frac{1}{2}MR^2)\omega_o  = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega

now we have

\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}

\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}

6 0
3 years ago
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